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Thanks Mike. I have just finished double checking everything, physically. No problems found so far. I am not able to detect the oscillator at the crystal frequency. I think this, for whatever

C6/C7 is ok. I double checked each of these with a LC Meter IIB as I was putting them in. They both are labeled 101J. C12/C13 are correct also. I double checked the windings of T1, and it looks

Cecil, I just had a look at the powerSDR-SR40 code. Looking at sdr.c line, I can see that the signal is indeed at 11025 Hz. The RX oscillator runs at -11025 hz. Now, does that mean, the spectrum we

Frank, Looking at the picture: http://www.amqrp.org/kits/softrock40/SoftRock40_v4.0.gif I couldn't understand how the mixing down to the magic number 11250 is obtained.. Would you mind explaining

The SoftRock itself doesn't convert all input signal to 11,250Hz, it has a fixed crystal oscillator, were the signal ends up depends where in the pass-band the original signal is. Some will be at 1KHz

Thanks,,, very fun following the list. I'm not likely to get involved with this type project at the hardware level, my current interests are mostly tube gear, but current interests do change over

If you want to run from a 6 volt wall wart you should consider the changes by N1RX to substitute a 7805 for Q1 (see http://groups.yahoo.com/group/softrock40/message/114). I have used a USB extension

Computers can be quirky! Try using the microphone input on your computer instead of the line input just be be sure thats not what its trying to get input from. If you have a scope, feed a signal into

wrote: There's a decent description of the Weaver method in the DSP section of all the recent ARRL Handbooks. The SoftRock itself does no demodulation. All it does is sample the incoming signal in

I forgot to add that I rewound T1 last evening thinking that might have been a problem. I wound the bifilar with green wire this time, that I had left over from the K2 kit. I could identify the

John, I do not hear a carrier at any of those freqs. I double checked my method by sniffing the XG1 and I hear it at 7.040. What level would the clock frequencies be at? I just have a peice of

Hi Stan, the S9 carrier at 7.056 is normal (even without the device powered). If you have a scope you can should check the clock signals that are shown on the schematic (QSD Timing Diagram) and also

I finished the SoftRock last night and I plugged it in to one of my USB hubs...no green light from the LED. I put it back on the bench and verified everything and all looked ok. After messing with it

Oh, and remember: Shannon's Theorem only addresses bandwidth, not absolute frequency. Having the signal sampled in quadrature means the phase of the signal is disambiguated, since the I and Q parts

Welcome to the club :-) Yes to all. display.... Here's the deal. Shannon's Theorem tells you that you need 2N bits/sec to capture N bits of bandwidth. If you're only sampling the real part, ie in

Normally yes, to sample 48KHz signals you need 96KHz sampling rate, but the I and Q are quadrature signals and are only at 24 KHz each. 7036KHz 20 KHz signal 7046KHz 10 KHz signal 7056KHz 0 KHz

WOW, thanks Tony. It's good news for me. I have been designing a crystal oscillator coupled to a divide by N counter. I had assumed no more than 24 Khz (12 KHz on each side of the center

Hi Art, Yes to all for questions below. Both I and Q channels are being individually sampled at 48 kHz which provides for the 48 kHz of bandwidth. 73, Tony KB9YIG

OK, now I'm really confused........... One question.......... If I fire up my softrock-40 on my computer that has the ability to sample at 48 Khz (with the center frequency at 7056 KHz).... Will I

The PC is able to "put" a phoney carrier anywhere in that 48KHz space so that the signal is demodulated correctly into audio. This is the answer I wanted to know for the POwerSDR sotfware.